Probability with Permutations and Combinations

1. GEOMETRY Five students are asked to randomlyselect and name a polygon from the group shownbelow. What is the probability that the first twostudents choose the triangle and quadrilateral, in thatorder?

SOLUTION:

Find the total number of outcomes. There are 5polygons. Therefore, 5 polygons can be chosen in 5!or 120 ways. Find the number of favorable outcomes. This is thenumber of permutations of the other polygons if thetriangle is chosen first and the quadrilateral is chosensecond: (5 – 2)! or 3!. Calculate the probability.

The probability that the first two students choose the

triangle and then the quadrilateral is .

ANSWER:

2. PLAYS A high school performs a production of ARaisin in the Sun with each freshman English classof 18 students. If the three members of the crew aredecided at random, what is the probability that Chaseis selected for lighting, Jaden is selected for props,and Emelina for spotlighting?

SOLUTION:

Since choosing students for the particular roles is away of ranking the members, order in this situation isimportant. The number of possible outcomes in the samplespace is the number of permutations of 18 peopletaken 3 at a time, 15P3.

Among these, there is only one particulararrangement in which Chase is selected for lighting,Jaden is selected for props, and Emelina for

spotlighting. Therefore, the probability is

ANSWER:

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12-3 Probability with Permutations and Combinations

3. DRIVING What is the probability that a licenseplate using the letters C, F, and F and numbers 3, 3, 3,and 1 will be CFF3133?

SOLUTION:

There are 3 letters and 4 digits This is a permutationwith repetition. There are a total of 7 objects in which F appearstwice and 3 appears three times. The number of distinguishable permutations is

The total number of possible outcomes is 420 andthere is only one favorable outcome which isCFF3133. Therefore, the probability is

ANSWER:

4. CHEMISTRY In chemistry lab, you need to test sixsamples that are randomly arranged on a circulartray.

a. What is the probability that the arrangement shownis produced?b. What is the probability that test tube 2 will be inthe top middle position?

SOLUTION:

a. The number of distinguishable permutations of nobjects arranged in a circle with no fixed reference

point is The 6 samples can be arranged

in 5! = 120 different ways. There is only onefavorable outcome as shown, so the probability is

b. Because the samples are arranged with a fixedreference point (the top middle position), this is alinear permutation. So, there are 6! or 720 ways inwhich the samples can be arranged. The number offavorable outcomes is the number of permutations ofthe other 5 samples given that test tube 2 will be inthe top middle position, 5! or 120. Therefore, the

probability is

ANSWER:

a.

b.



5. Five hundred boys, including Josh and Sokka, entereda drawing for two football game tickets. What is theprobability that the tickets were won by Josh andSokka?

SOLUTION: Two students are chosen out of 500. The order inwhich the two students are chosen does not matter,so the total number of outcomes is:

There is only 1 favorable outcome, so the probability

is .

ANSWER:

6. CONCERTS Nia and Chad are going to a concertwith their high school’s key club. If they choose aseat on the row below at random, what is theprobability that Chad will be in seat C11 and Nia willbe in C12?

SOLUTION:

Since choosing seats is a way of arranging thestudents, order in this situation is important. There are 12 seats. The number of possible outcomesin the sample space is the number of permutations of12 seats taken 2 at a time, 12P2.

Among these, there is only one particulararrangement in which Chad will be in seat C11 and

Nia will be in C12. Therefore, the probability is

ANSWER:



7. FAIRS Alfonso and Colin each bought one raffleticket at the state fair. If 50 tickets were randomlysold, what is the probability that Alfonso got ticket 14and Colin got ticket 23?

SOLUTION:

Since buying tickets is a way of arranging the people,order in this situation is important. There are 50 tickets. The number of possibleoutcomes in the sample space is the number ofpermutations of 50 tickets taken 2 at a time, 50P2.

Among these, there is only one particulararrangement in which Alfonso gets ticket 14 andColin gets ticket 23. Therefore, the probability is

ANSWER:

8. MODELING The table shows the finalists for afloor exercises competition. The order in which theywill perform will be chosen randomly.

a. What is the probability that Cecilia, Annie, andKimi are the first three gymnasts to perform, in anyorder?b. What is the probability that Cecilia is first, Annie is

second, and Kimi is third?

SOLUTION:

a. The number of possible outcomes is the number ofarrangements of 7 performers taken 7 at a time. So,the number of possible outcomes is 7! = 5040. The first 3 finalists can be arranged in 3! = 6 ways.The rest of the finalists can be arranged 4! ways. Therefore, the number of favorable outcomes is (4!)(3!) = (24)(6) = 144. So, the probability is

b. The number of possible outcomes is 5040. Thenumber of favorable outcomes with Cecilia first,Annie second, and Kimi third is 1. After the firstthree places are set, the rest of the finalists can bearranged 4! ways. Therefore, the probability is

ANSWER:

a.

b.



9. JOBS A store randomly assigns their employeeswork identification numbers to track productivity.Each number consists of 5 digits ranging from 1–9. Ifthe digits cannot repeat, find the probability that arandomly generated number is 25,938.

SOLUTION:

The number of possible outcomes is the number ofarrangements of 9 digits taken 5 at a time, 9P5.

The only favorable outcome is the arrangement25,398, so the probability is

ANSWER:

10. GROUPS Two people are chosen randomly from agroup of ten. What is the probability that Jimmy wasselected first and George second?

SOLUTION:

Because choosing people is a way of ranking themembers, order in this situation is important. The number of possible outcomes in the samplespace is the number of permutations of 10 peopletaken 2 at a time, 10P2.

Among these, there is only one particulararrangement in which Jimmy is selected first and

George second. Therefore, the probability is

ANSWER:

11. MAGNETS Santiago bought some letter magnetsthat he can arrange to form words on his fridge. If herandomly selected a permutation of the letters shownbelow, what is the probability that they would formthe word BASKETBALL?

SOLUTION:

There are 10 letters in which A, B, and L eachappear twice. The number of distinguishable permutations is

The total number of possible outcomes is 453,600 andthere is only one favorable outcome which isBASKETBALL. Therefore, the probability is

ANSWER:



12. ZIP CODES What is the probability that a zip coderandomly generated from among the digits 3, 7, 3, 9,5, 7, 2, and 3 is the number 39372?

SOLUTION:

There are 8 digits in which 3 appears three times and7 appears twice. The number of distinguishable permutations is

The total number of possible outcomes is 3360 andthere is only one favorable outcome which is 39372.Therefore, the probability is

ANSWER:

13. GROUPS Keith is randomly arranging desks intocircles for group activities. If there are 7 desks in hiscircle, what is the probability that Keith will be in thedesk closest to the door?

SOLUTION:

Because the people are seated with a fixed referencepoint, this is a linear permutation. So, there are 7! or 5040 ways in which the peoplecan be seated. The number of favorable outcomes isthe number of permutations of the other 6 peoplegiven that Keith will be in the desk closest to the door,

6! or 720. Therefore, the probability is

ANSWER:

14. AMUSEMENT PARKS Sylvie is at an amusementpark with her friends. They go on a ride that hasbucket seats in a circle. If there are 8 seats, what isthe probability that Sylvie will be in the seat farthestfrom the entrance to the ride?

SOLUTION:

Because the people are seated with a fixed referencepoint, this is a linear permutation. So, there are 8! or 40,320 ways in which the peoplecan be seated. The number of favorable outcomes isthe number of permutations of the other 7 peoplegiven that Sylvie will be in the seat farthest from theentrance to the ride, 7! or 5040. Therefore, theprobability is

ANSWER:



15. PHOTOGRAPHY If you are randomly placing 24photos in a photo album and you can place fourphotos on the first page, what is the probability thatyou choose the photos shown?

SOLUTION:

You are choosing 4 photos from a set of 24 photos,and the 4 chosen photos can be placed in any order.So, the number of possible outcomes is 24C4.

There is only one favorable outcome, so theprobability is

ANSWER:

16. ROAD TRIPS Rita is going on a road trip acrossthe United States. She needs to choose from 15 citieswhere she will stay for one night. If she randomlypulls 3 city brochures from a pile of 15, what is theprobability that she chooses Austin, Cheyenne, andSavannah?

SOLUTION:

You are choosing 3 cities from a set of 15 cities, andorder is not important. So, the number of possibleoutcomes is 15C3.

The number of favorable outcomes is only one, whichis choosing the cities Austin, Cheyenne, and

Savannah. Therefore, the probability is

ANSWER:

17. STRUCTURE Use the figure below. Assume thatthe balls are aligned at random.

a. What is the probability that in a row of 8 pool balls,the solid 2 and striped 11 would be first and secondfrom the left?b. What is the probability that if the 8 pool balls weremixed up at random, they would end up in the ordershown?c. What is the probability that in a row of seven balls,with three 8-balls, three 9-balls, and one 6-ball, thethree 8-balls would be to the left of the 6-ball and thethree 9-balls would be on the right?d. If the eight original balls were randomlyrearranged and formed a circle, what is theprobability that the 6-ball is next to the 7-ball?

SOLUTION:



a. The number of possible outcomes in the samplespace is the number of permutations of 8 balls taken8 at a time, 8! = 40,320. The number of favorable outcomes is the number ofpermutations of the remaining 6 balls after fixing thesolid 2 and striped 11 at the first and second from theleft, 6! = 720.

Therefore, the probability is

b. The number of possible outcomes is 40,320 and thenumber of favorable outcomes is the only one as

shown. Therefore, the probability is

c. The required probability is the probability of gettingan arrangement of 8886999 from three 8-balls, three9-balls and one 6-ball. So, the number ofdistinguishable permutations is

There is only one favorable outcome, which is

8886999, so the probability is

d. Eight balls are arranged in a circle with one fixedball (7 ball). So, the circular permutation of arranging8 balls with 1 fixed ball is 7!. The 6-ball can be arranged either to the left of ball 7or right of ball 7. So, the circular permutation ofarranging 8 balls with 2 fixed balls is .

ANSWER:

a.

b.

c.

d.

18. How many lines are determined by 10 randomlyselected points, no 3 of which are collinear? Explainyour calculation.

SOLUTION:

Two points determine a line. If no 3 points arecollinear, then each line will consist of exactly 2points. Because there are 10 total points, the numberof lines is the combination of 10 points taken 2 at a

time, which is or 45.

ANSWER:

45; Sample answer: The number of lines is thecombination of 10 objects taken 2 at a time, which is

or 45.



19. Suppose 7 points on a circle are chosen at random, asshown.

a. Using the letters A through E, how many ways canthe points on the circle be named?b. If one point on the circle is fixed, how manyarrangements are possible?

SOLUTION:

a. The number of distinguishable permutations of nobjects arranged in a circle with no fixed reference

point is

The 7 points can be named in a circular way in 6! =720 different ways. b. Because the points are named with a fixedreference point, this is a linear permutation. There are7! or 5040 ways in which the points can be named.

ANSWER:

a. 720b. 5040

20. RIDES A carousel has 7 horses and one bench seatthat will hold two people. One of the horses does notmove up or down.

a. How many ways can the seats on the carousel berandomly filled by 9 people?

b. If the carousel is filled randomly, what is theprobability that you and your friend will end up in thebench seat?c. If 6 of the 9 people randomly filling the carouselare under the age of 8, what is the probability that aperson under the age of 8 will end up on the onehorse that does not move up or down?

SOLUTION:

a. The number of ways in which the 9 seats can befilled is the permutation of 9 people taken 9 at a time.So, the number of ways is 9! = 362,880. b. The total number of possible outcomes is 362,880.Because the carousel is filled with a fixed referencepoint, this is a linear permutation. So, there are 7! or 5040 ways to arrange theremaining 7 people after you and your friend areplaced in the bench seat. Also, for each arrangementthere is an alternate arrangement by interchangingthe position of you and your friend. Therefore, theprobability is

c. The total number of arrangements of the 9 peopleis or 504. The number of favorable outcomes is

the number of distinguishable permutations of theother eight places if a child under 8 is on the horse

that does not move up or down: or 336. Calculate the probability.

ANSWER:

a. 362,880

b.

c.



21. LICENSES A camera positioned above a trafficlight photographs cars that fail to stop at a red light.In one unclear photograph, an officer could see thatthe first letter of the license plate was a Q, thesecond letter was an M or an N, and the third letterwas a B, P, or D. The first number was a 0, but thelast two numbers were blurry. How many possiblelicense plates fit this description?

SOLUTION:

By the Fundamental Counting Principle, the numberof possible outcomes in a sample space can be foundby multiplying the number of possible outcomes fromeach stage or event. The first letter of the license plate is Q. There are 2 choices for the second letter.There are 3 choices for the third letter. The first number is 0.There are 10 choices for the second digit.There are 10 choices for the third digit. Therefore, the total number of choices is 1 × 2 × 3 ×1 × 10 × 10 = 600.

ANSWER:

600

22. MULTI-STEP Corie is interested in playing a statelottery game in which she needs to choose a certainthree-digit number to win. The different gameoptions are shown in the table. Which option shouldshe choose? Explain your solution process.

SOLUTION: First, find the probability of winning with each option.There are 10³ or 1000 three-digit numbers, so thereare 1000 possible choices.

Straight: She has only one chance of winning:P(straight) = 1 ÷ 1000 or 0.001.

3-Way Boxed: There are three winning numbers:P(3-Way) = 3 ÷ 1000 or 0.003.

6-Way Boxed: There are six winning numbers: P(6-Way) = 6 ÷ 1000 or 0.006.

Straight-Box: There are still 6 winningcombinations. P(Straight-Box) = 6 ÷ 1000 or 0.006. Because the odds of winning are the same for a 6-Way Boxed and a Straight-Box, Corie should chooseeither of these options to play.

ANSWER: First, find the probability of winning with each option.There are 10³ or 1000 three-digit numbers, so thereare 1000 possible choices.

Straight: She has only one chance of winning:P(straight) = 1 ÷ 1000 or 0.001.

3-Way Boxed: There are three winning numbers:P(3-Way) = 3 ÷ 1000 or 0.003.

6-Way Boxed: There are six winning numbers: P(6-Way) = 6 ÷ 1000 or 0.006.

Straight-Box: There are still 6 winningcombinations. P(Straight-Box) = 6 ÷ 1000 or 0.006. Because the odds of winning are the same for a 6-Way Boxed and a Straight-Box, Corie should chooseeither of these options to play.



23. MODELING Fifteen boys and fifteen girls entereda drawing for four free movie tickets. What is theprobability that all four tickets were won by girls?

SOLUTION:

Four people can be chosen from 30 people in 30C4ways and 4 girls can be chosen from 15 girls in 15C4ways.

The number of possible outcomes is 27,405 and thenumber of favorable outcomes is 1365. Therefore,

the probability is

ANSWER:

24. REASONING A student claimed that permutationsand combinations were related by . Usealgebra to show that this is true. Then explain why

and differ by the factor r!.

SOLUTION:

Sample answer:

nCr and nPr differ by the factor r! because there arealways r! ways to order the groups that are selected.Therefore, there are r! permutations of eachcombination.

ANSWER:

Sample answer:

nCr and nPr differ by the factor r! because there arealways r! ways to order the groups that are selected.Therefore, there are r! permutations of eachcombination.



25. OPEN-ENDED Describe a situation in which theprobability is given by .

SOLUTION:

In this situation, you have 7 objects taken 3 at a timeand order is not important because it is a combination. Sample answer: A bag contains seven marbles thatare red, orange, yellow, green, blue, purple, and black.The probability that the orange, blue, and blackmarbles will be chosen if three marbles are drawn atrandom can be calculated using a combination.

ANSWER:

Sample answer: A bag contains seven marbles thatare red, orange, yellow, green, blue, purple, and black.The probability that the orange, blue, and blackmarbles will be chosen if three marbles are drawn atrandom can be calculated using a combination.

26. CONSTRUCT ARGUMENTS Is the followingstatement sometimes, always, or never true?Explain.

nPr = nCr

SOLUTION:

r! = 1 when r = 0 or 1. However, objects cannot betaken 0 at a time. The statement is true when r is 1. Therefore, thestatement is sometimes true.

ANSWER:

Sometimes; sample answer: The statement is truewhen r is 1.



27. PROOF Prove that .

SOLUTION:

Set the two sides equal to each other. Develop eachside using the formulas in this lesson and simplify.Working with the left side will eventually reduce it tothe right side.

ANSWER:

28. WRITING IN MATH Compare and contrastpermutations and combinations.

SOLUTION:

Sample answer: Both permutations and combinationsare used to find the number of possible arrangementsof a group of objects. The order of the objects isimportant in permutations but not in combinations.

ANSWER:

Sample answer: Both permutations and combinationsare used to find the number of possible arrangementsof a group of objects. The order of the objects isimportant in permutations but not in combinations.

29. Steven, Yasmin, Tyler, Vanessa, and Zack have beennominated to be club officers (president and vicepresident). Assuming the officers are chosen atrandom from among the five nominees, what is theprobability that Steven is chosen to be the presidentand Yasmin is chosen to be the vice president? A B C D

SOLUTION: Step 1: Because choosing officers is a way ofranking team members, order in this situation isimportant. The number of possible outcomes in the samplespace is the number of permutations of 5 peopletaken 2 at a time, which is .

Step 2: The number of favorable outcomes is thenumber of permutations of the 2 students in theirspecific positions. This is 1!, or 1. Step 3: The probability of Steven being chosen as

the president and Yasmin as the vice president is . So, the correct answer is choice D.

ANSWER: D



30. Manuel must choose two books from a list of sevento read for book reports in his literature class. Howmany different pairs of books could he choose?

A 10B 21C 42D 5040

SOLUTION: Because the order in which the books are chosendoes not matter, the number of possible outcomes inthe sample space is the number of combinations of 7books taken 2 at a time.

So, the correct answer is choice B.

ANSWER: B

31. Every student at Shellie’s school is assigned a 4-digitPIN to access the school’s computers. The first digitof the code is 0 or 1. The remaining digits are chosenfrom the numbers 2 through 9, and no number may beused more than once in a code. What is theprobability that Shellie is assigned the PIN 1234? A B C

D E

SOLUTION: Step 1: First, consider how many choices Shellie hasfor each digit in her PIN: For the first digit, there are 2 numbers to pick from, 0

or 1. So, this is . For the last three digits, there are 8 numbers to pickfrom, 2 through 9. So, this is

.

So, there are 672 possible PINs.

Step 2: There is only 1 favorable arrangement, 1234.

Step 3: The probability that Shellie is assigned PIN

1234 = . So, the correct answer is choice D.

ANSWER: D



32. The coach is going to select a team of 5 from among10 players. Find the probability that John, a particularplayer, is on the team.

SOLUTION: Because order does not matter, the number ofpossible outcomes in the sample space is the numberof combinations of 10 people taken 5 at a time, .

Because John must occupy one of the places on theteam, you then need to determine .

So, the probability that John will be on the team is

.

ANSWER:

33. Marsala inserts three $5-bills into 3 of 10 envelopes.The other 7 she leaves empty. She has each of her 10soccer teammates choose one of the envelopes. Howmany different ways can the money be won by herteammates?

A 120B 35C 720D 3,628,800

SOLUTION: Because the order of the envelopes with the $5 inthem does not matter, the number of possibleoutcomes in the sample space is the number ofcombinations of 10 envelopes taken 3 at a time, .

So, the correct answer is choice A.

ANSWER: A



34. Midori has tiles with the letters J, K, L, M, N, P, andQ. She chooses three of the tiles at random. What isthe probability that the tiles name three collinearpoints on the coordinate plane shown here?

A

B

C

D

SOLUTION: Because the order does not matter, the number ofpossible outcomes in the sample space is the numberof combinations of 7 letters taken 3 at a time, .

There are only 2 favorable outcomes, collinear pointsJ, K, L or collinear points L, M, N, so the probabilityis . The correct answer is choice C.

ANSWER: C

35. MULTI-STEP Assuming that any arrangement ofletters forms a ‘word’, what is the probability that a‘word’ formed from the letters of the wordSQUARE: a. starts with E?b. starts with E and ends with A?

SOLUTION: a. Order is important in this situation, so the numberof possible outcomes in the sample space is thenumber of permutations of 6 letters taken 1 at a time,

.

The number of favorable outcomes is 1, which is theletter E, so the probability is . b. Given that the probability of the word starting withE is , we must now consider the remainder of theword. Order is important, so the number of possibleoutcomes in the new sample space is the number ofpermutations of 5 letters taken 1 at a time, .

Again, the number of favorable outcomes is 1, which

is the letter A, so the probability is . To find the probability of both events, multiply.

ANSWER:

a.

b.



36. 15 student council members select a committee of 3for a project, and then select one of the three to bethe liaison for the project. a. In how many ways is this possible?b. What is the probability that Carmen is in thecommittee?c. What is the probability that Carmen is the liaisonfor the project?

SOLUTION: a. Because the order does not matter, the number ofpossible outcomes in the sample space is the numberof combinations of 15 members taken 3 at a time,

.

Because there are 3 possible favorable outcomes,multiply 455 × 3, or 1365. b. The probability that Carmen is in the committee is

. c. To determine the probability that Carmen is theliaison, multiply .

ANSWER: a. 1365b.

c.



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Probability with Permutations and Combinations